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#combustion

24 public questions tagged with this topic.

CHEMISTRY

A compound burns in oxygen to produce 8.8 g of CO₂ and 3.6 g of H₂O. What is its empirical formula? (Atomic masses: C =

Given: A compound burns in oxygen to produce 8.8 g of CO₂ and 3.6 g of H₂O. What is its empirical formula? (Atomic masses: C = 12, H = 1, O = 16) These values define the system as per NCERT data. Formula: Mass of C = (12 / 44) × 8.8 = 2.4 g. This is standard NCERT relation. Substitution & Calculation: Mass of H = (2 / 18) × 3.6 = 0.4 g. Moles: C = 2.4 / 12 = 0.2, H = 0.4 / 1 = 0.4. Ratio: 0.2 / 0.2 : 0.4 / 0.2 = 1 : 2. Empirical formula = CH₂. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

A hydrocarbon on combustion yields 17.6 g of CO₂ and 7.2 g of H₂O. What is its empirical formula? (Atomic masses: C = 12

Given: A hydrocarbon on combustion yields 17.6 g of CO₂ and 7.2 g of H₂O. What is its empirical formula? (Atomic masses: C = 12, H = 1, O = 16) Formula: Mass of C = (12 / 44) × 17.6 = 4.8 g. Substitution & Calculation: Mass of H = (2 / 18) × 7.2 = 0.8 g. Moles: C = 4.8 / 12 = 0.4, H = 0.8 / 1 = 0.8. Ratio: 0.4 / 0.4 : 0.8 / 0.4 = 1 : 2. Empirical formula = CH₂. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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Fundamental Concepts - Reaction Mechanism - Fission and Reaction Intermediates

A hydrocarbon with 9 σ bonds produces 5 moles of CO₂ per mole upon combustion. What is its molecular formula?

5 carbons (from 5 CO₂). CH₃CH₂CH₂CH₂CH₃ (pentane) has 9 σ (8 C-H, 1 C-C from single bonds), no π, and fits C₅H₁₂.

Ref: NCERT Class 11 Chemistry > Chapter 8: Organic Chemistry - Some Basic Principles and Techniques > Topic: Fundamental Concepts - Reaction Mechanism - Fission and Reaction Intermediates

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Stoichiometry and Stoichiometric Calculations and Limiting Reagent

What volume of CO₂ at STP is produced when 10 g of C₄H₁₀ is burned completely? (Molar mass: C₄H₁₀ = 58 g/mol)

Reaction: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O. Moles of C₄H₁₀ ≈ 0.1724 mol. 2 mol produce 8 mol CO₂; 0.1724 mol produce ≈ 0.6896 mol. Volume = 0.6896 × 22.4 ≈ 15.45 L.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Stoichiometry and Stoichiometric Calculations and Limiting Reagent

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Stoichiometry and Stoichiometric Calculations and Limiting Reagent

What volume of CO₂ at STP is produced when 15 g of C₃H₆ is burned completely? (Molar mass: C₃H₆ = 42 g/mol)

Reaction: 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O. Moles of C₃H₆ = 15/42 ≈ 0.3571 mol. 2 mol C₃H₆ produce 6 mol CO₂; 0.3571 mol produce ≈ 1.0714 mol. Volume = 1.0714 × 22.4 ≈ 24 L.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Stoichiometry and Stoichiometric Calculations and Limiting Reagent

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Reactions in Solutions and Mass Percentage and Numericals

What volume of O₂ at STP is required to burn 7.8 g of C₃H₆ completely to CO₂ and H₂O? (Molar mass: C₃H₆ = 42 g/mol)

Reaction: 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O. Moles of C₃H₆ ≈ 0.1857 mol. 2 mol need 9 mol O₂; 0.1857 mol need ≈ 0.8357 mol. Volume = 0.8357 × 22.4 ≈ 18.72 L.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Reactions in Solutions and Mass Percentage and Numericals

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